Algebra and Functions
Algebra and Functions carries a large share of the pure papers: polynomial manipulation, the factor and remainder theorems, partial fractions, modulus and inequality work, and the behaviour of rational functions (asymptotes, domains and ranges). Questions usually combine two of these, so fluency with algebraic form matters more than memorised tricks.
Polynomials: factor and remainder theorems
- Dividing f(x) by (x − a) leaves remainder f(a); the division is exact if and only if f(a) = 0.
- Dividing by (ax + b) leaves remainder f(−b/a) - substitute the value that makes the divisor zero.
- Synthetic division gives the quotient, whose degree is one lower than f(x).
- A repeated factor (x − a)² requires f(a) = 0 and f′(a) = 0.
Partial fractions
- Reduce to a proper fraction first: if the degree of the numerator is at least that of the denominator, divide.
- Distinct linear factors: A/(x − a) + B/(x − b).
- Repeated linear factor: A/(x − a) + B/(x − a)².
- Irreducible quadratic: (Bx + C)/(x² + px + q).
- Choose convenient x values (cover-up) instead of expanding and comparing coefficients.
Inequalities and modulus
- Solve f(x) < g(x) with a sign table for f − g; never multiply across by an expression of unknown sign.
- |x − a| < k means a − k < x < a + k; |x − a| > k splits into two rays.
- |f(x)| = g(x) requires g(x) ≥ 0 and solving f = ±g.
Worked example
Express (3x + 5)/((x − 1)(x + 2)) in partial fractions. Write 3x + 5 = A(x + 2) + B(x − 1). Put x = 1: 8 = 3A, so A = 8/3. Put x = −2: −1 = −3B, so B = 1/3. The expression is 8/(3(x − 1)) + 1/(3(x + 2)). Check at x = 0: left side is 5/(−2) = −2.5, right side is −8/3 + 1/6 = −2.5. Confirmed.
Traps
- Substituting x = a when the divisor is (ax + b) rather than x = −b/a.
- Skipping the division step when the fraction is improper.
- Squaring both sides of a modulus equation and keeping the extraneous root.